If y -1- x 1- x 21- x 4 - - -1- x 2 n - then dy - dx x -0-A- 0B- 1 - 2C- 1D- 2

The correct option is C 1 y=1 x^4n/1 x (dy/dx )_x=0=1 ...

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Properties of Iota

If y =1+ x 1+...

Question

If $y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . . . (1 + x^{2 n}), then {(\frac{d y}{d x})}_{x = 0} =$

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Similar questions

y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . . (1 + x^{2^{n}})

{(\frac{d y}{d x})}_{x = 0} =

If $y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . . . (1 + x^{2 n}), then {(\frac{d y}{d x})}_{x = 0} =$

y = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}})

\frac{d y}{d x}

y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . . . . . . . (1 + x^{2 n})

(\frac{d y}{d x})

x = 0

y = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}})

\frac{d y}{d x}

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